Problem: Find the value(s) of $x$ such that $8xy-12y+2x-3=0$ is true for all values of $y$.
Solution: The given equation can be factored as $$
0=8xy-12y+2x-3=4y(2x-3)+(2x-3)=(4y+1)(2x-3).
$$For this equation to be true for all values of $y$ we must have $2x-3=0$, that is,  $x=\boxed{\frac{3}{2}}$.